Stuck on radicals...please help!

**coopsterdude** · May 14th 2006, 07:41 AM

Hi, I’ve been stuck on this problem for a week and can’t seem to find any example that would help me to solve it. I would really appreciate it if someone would explain the steps to solving this problem. Thank you so much.

**ThePerfectHacker** · May 14th 2006, 06:48 PM

I developed a theorem (imcomplete) sometime ago, that if you have an integral linear combination of irrational distinct radicals equal to each other then there corresponding parts are equal.
For example, if
$1+\sqrt{2}+\sqrt{3}=x+y\sqrt{2}+z\sqrt{3}$
where $x,y,z$ are integers then,
$x=1,y=1,z=1$ .
Now if you have,
$1+\sqrt{2}+\sqrt{3}=x+y\sqrt{5}$
where $x,y$ are integers is just impossible,
If you have that,
$\sqrt{16-6\sqrt{3}}=x+y\sqrt{3}$
where $x,y$ are integers then,
$16-6\sqrt{3}=(x+y\sqrt{3})^2$
Thus,
$16-6\sqrt{3}=(x^2+3y^2)+2xy\sqrt{3}$
Then,
$\left\{ \begin{array}{c}x^2+3y^2=16\\ 2xy=6$ is just impossible in integers.

This is my reasoning that why I would not expect a simpler solution with the integers.

**coopsterdude** · May 15th 2006, 09:21 AM

Thanks for all your help and I'm am so sorry about breaking all those rules.

**CaptainBlack** · May 15th 2006, 09:50 AM

Originally Posted by coopsterdude

Hi, I’ve been stuck on this problem for a week and can’t seem to find any example that would help me to solve it. I would really appreciate it if someone would explain the steps to solving this problem. Thank you so much.

How about:

$<br /> \sqrt{12-6\sqrt{3}}=3-\sqrt{3}<br />$

(using the convention that $\sqrt{}(\cdot)$ denotes the positive
root.)

Just goes to show that it never does to be too dogmatic PH

RonL

**ThePerfectHacker** · May 15th 2006, 10:23 AM

Originally Posted by CaptainBlack

How about:

Just goes to show that it never does to be too dogmatic PH

No I did not make a mistake in what I said.
Note I made a mistake in doing
$\sqrt{16-6\sqrt{3}}$ the wrong problem.
If you used my method then you will see that it works

-------------------------
I base it on the fact that,
$a_0+a_1\sqrt{b_1}+...+a_n\sqrt{b_n}$
is always irrational whenver all are integers and the b_i are all non-squares and a_i are not zero.

Then given
$a_0+a_1\sqrt{b_1}+...+a_n\sqrt{b_n}$ = $c_0+c_1\sqrt{b_1}+...+c_n\sqrt{b_n}$
You can write,
$d_0+d_1\sqrt{b_1}+...+d_n\sqrt{b_n}=0$ where $d_i=a_i-c_i$
Which is rational an impossibility unless a_i=c_i giving zeros.

**Ranger SVO** · May 16th 2006, 05:48 PM

Originally Posted by CaptainBlack

How about:

$<br /> \sqrt{12-6\sqrt{3}}=3-\sqrt{3}<br />$

(using the convention that $\sqrt{}(\cdot)$ denotes the positive
root.)

Just goes to show that it never does to be too dogmatic PH

RonL

I would really like to see the steps to get to the simplified form. Thank you for your time.

**ThePerfectHacker** · May 16th 2006, 06:38 PM

Originally Posted by Ranger SVO

I would really like to see the steps to get to the simplified form. Thank you for your time.

You can use my method which I shown, (I only did the wrong example which is why it did not work).
You attempt to find integers $x,y$ such as,
$\sqrt{12-6\sqrt{3}}=x+y\sqrt{3}$
Thus,
$12-6\sqrt{3}=(x+y\sqrt{3})^2$
Thus,
$12-6\sqrt{3}=(x^2+3y^2)+2xy\sqrt{3}$
Then, by my previous post,
$\left\{ \begin{array}{c}12=x^2+3y^2\\ -6=2xy$
Look at the second equation,
$-6=2xy$
This, is equivalent to
$xy=-3$
Since $x,y$ are integers the only possibilities are,
$(x,y)=(-1,3),(3,-1),(-3,1),(1,-3)$
Now we check which one of these satisfies the second equation,
we see that,
$\begin{array}{c} (3)^2+3(-1)^2=12\\ (-3)^2+3(1)^2=12$
Thus,
$3-\sqrt{3}$
$\sqrt{3}-3$
Both of these are true, indeed,
$(\sqrt{3}-3)^2=(3-\sqrt{3})^2=12-6\sqrt{3}$
But we need to chose,
$3-\sqrt{3}$
Because square roots are non-negative and,
$\sqrt{3}-3<0$

**Ranger SVO** · May 17th 2006, 05:49 AM

Thank you, I understand it now but I wonder if a High School Math Teacher would have intended a simpler solution.

I am finished with finals and had time to look at the problem.

Thread: Stuck on radicals...please help!

LinkBack

Thread Tools

Display

Stuck on radicals...please help!

Similar Math Help Forum Discussions

solving equations w/ radicals and exponents problem, stuck

help with radicals

Radicals

Radicals?

Stuck, Stuck, Stuck - Need Help Urgently

Search Tags